What is a Phasor Diagram in AC circuit Analysis: Phasor Algebra

An alternating current waveform having a specific frequency, when comparing it with another AC quantity (AC voltage, Current, power etc.), an effect of leading or lagging can be seen in time domain waveform. The time domain representation of different waveforms can be more intuitive but tedious in case multiple AC quantities and more data operation. To work with a large amount of AC parameters waveforms, a simple representation can be introduced to differentiate among their magnitude and phase angle is called Phasor Representation of AC circuit.

Phasor representation of an AC quantity is a counter-clockwise rotating vector originating from the origin of XY-axes (zero point) in a specific direction. The direction of phasor shows the angle of a sinusoidal waveform when it passes through the vertical axis.

The magnitude of the phasor represents the magnitude of the quantity and the angle with the X-axis show the phase angle. The phasors are used while comparing and performing mathematical operations on AC quantities. Remember that phasors are used only when the frequency is the same.

Difference between Phasor and vector :

Before going further let me clear the confusion between vector and phasor. Vectors are stationary making a certain angle with the real axis. Where phasor is a radius-vector of a circle which rotates anti-clockwise. The phasor direction shows the initial position of the rotation. If the only vertical component of a phasor is drawn against angle or time, you will get sinewave and for horizontal component, you will get cosine wave.

Phasor contains information about magnitude, phase angle, and angular frequency. Where vector tells us about magnitude and direction rather than phase angle. And vector is a stationary and broad term for directional quantities. In contrast, a phasor is a specific term referring to anti-clockwise rotational vectors.

Need of phasor diagram in AC Circuit Analysis :

While comparing two different waveforms in ac AC circuit analysis i.e. current and voltage it is possible to draw them on a same set of axes and visually analyze the difference between them. The this could be a very tedious and lengthy process with limited accuracy.

In a similar way, while performing the operation of multiplication, addition and many more, it is possible to perform the operation point by point. Again a very lengthy, exhaustive and time-consuming process with less accuracy.

Relationship between phasor and a sinusoidal waveform :

Suppose a vector rotating counter-clockwise with an angular frequency of $\omega$ radians/seconds. If the vertical component (Y-part) of the vector is plotted step-by-step as the vector moves, the plot will a sinusoidal waveform.

For a unit circle, when the vector is making zero degrees with the x-axis,
the vertical component is zero and at 90 degrees the y-component is maximum. Then it starts to decrease toward 180 degree and y-component to zero. Then the negative half-cycle starts by going toward the 270 degrees and vertical component negative maximum. Finally, vector rotation complete at 360 degrees and once again the vertical part of the vector goes to zero. The vertical component is plotted against the rotation angle, which represents the sinusoidal waveform of frequency $\omega /2\pi$ Hertz.

Instead of starting the rotation from the zero degrees, a vector my start rotation somewhere else, let’s say 30 degrees. So, how would the waveform look like?

The waveform, in this case, would start from ½ instead of zero and after completing positive and negative half-cycle, it will reach ½ to complete 360-degree rotation. In the following diagram, the waveform B leads the waveform A by 30 degrees and the same is represented by the corresponding phasors. Both phasors are plotted over another axis conveying the same concept of phase relationship.

Where the equations for the waveforms are

$Waveform, A=A_{m}\sin (\omega t)$

Where the phasor equations are

$a(t)= Im\lbrack \dot{A}e^{j\omega t}\rbrack$

The $e^{j\omega t}$ is the term which rotates the vectors counter-clockwise with the angular frequency of $\omega$.

Phasor algebra :

Sometimes it is necessary to add, subtract, multiply and divide alternating waveforms. If there is no phase shift between them, they can be added and subtracted just like scalar numbers. For example, if we have to add two voltages 50v and 30v with no phase difference, their resultant voltage will be 50+30=80v.

So, how to deal with phase-shifted alternating waveforms phasors?

As there is phase difference, so their direction at any instance is not identical. Taking it into account, the resultant effect can be found using vector operation rather than scalar operation.

For example, in series AC circuit, there is a need of phasor addition of voltages which may have a phase difference. To determine the resultant phasor, the law of parallelogram or the head and tail rule of vector addition is used.

How to add phasors with parallelogram law?

Continuing with the above example waveforms, A and B and its phasors. The phasor A and phasor B form two adjacent sides of a parallelogram, where the diagonal combining them is the resultant phasor.

Another way of performing phasor addition is head and tail rule. In which, the tail of a phasor is joined with the head of another phasor. A phasor combining the tail of first and the head second phasor represent resultant phasor. Both methods have the same output.

Phasor Subtraction:

Phasor addition and phasor subtraction are very similar operations. In case of subtraction, the direction of the phasor which you want to subtract is reversed. Then the addition is performed obtain the resultant phasor.

Conversion between polar and complex number:

Complex Numbers and Phasors:

As phasors are rotating vectors and it is considered that all the phasor of
a circuit like current and voltage phasors has the same frequency. So, dropping the rotation component $e^{j\omega t}$, we only have vector remaining. Now, this vector can be converted into vertical and horizontal components using Euler formula. Finding the rectangular components of the phasor B.

$B_{p}=\vert B\vert \angle30^{o}$

Vertical component: $B_{y}=\dot{B}\sin (30)$

Horizontal component: $B_{x}=\dot{B}\cos (30)$

The phasor can also be represented by the combination of these X and Y components as follow

$B_{c}=B_{x}+jB_{y}$

The complex number offer simpler addition and subtraction while multiplication and division are easy in polar form. Conversion between them takes place through Euler formula.

:

Suppose two phasors A and B, after dropping of rotational term and conversion to the complex form, we want to add with each other.

$A_{c}=A_{x}+jA_{y}$

$B_{c}=B_{x}+jB_{y}$

The addition process for complex numbers is very simple, addition is performed between real and imaginary parts separately. The resultant
complex number from the above addition will be

$R_{c}=Ac+B_{c}$

$R_{c}=\lbrack A_{x}+jA_{y}\rbrack +\lbrack B_{x}+jB_{y}\rbrack$

Subtraction through complex numbers:

In case of subtraction, nothing else will change except the sign, the above example will look something like this

$R_{c}=Ac-B_{c}$

$R_{c}=\lbrack A_{x}+jA_{y}\rbrack -\lbrack B_{x}+jB_{y}\rbrack$

Polar Form:

Another way to express a vector in a 2-dimensional plane is Polar form, by telling the magnitude of the vector and the angle which it makes with the horizontal axis. Taking the above example, the phasor B after dropping the rotational term can be written as

$B_{p}=\vert B\vert \angle30^{o}$

The angle is always counted anti-clockwise starting in the first quadrant of the Cartesian plane.

Multiplication of phasors through the polar form:

The preferred method for multiplication of vector is polar form because of simplicity. The magnitude of the vectors is multiplied where the angle between
them is added. Suppose the following two vectors are converted to polar form, where we want to multiply them

$A_{p}=\vert A\vert \angle0^{o}$

$B_{p}=\vert B\vert \angle30^{o}$

The resultant vector will be

$R_{p}=A_{p}\times B_{p}$

$R_{p}=\lbrack \vert A\vert \angle0^{o}\rbrack \times \lbrack \vert B\vert \angle30^{o}\rbrack$

$R_{p}=\vert A\vert \vert B\vert \angle30^{o}$

Division of phasors through the polar form:

The division of phasor is a reciprocal process of multiplication. In polar form, the magnitude of the vector is divide where their angles are subtracted
. Continuing the above example phasors and their stationary vectors,
let me show you mathematically how to perform the division of phasors.

Suppose the above stationary vector in the polar form

$A_{p}=\vert A\vert \angle0^{o}$

$B_{p}=\vert B\vert \angle30^{o}$

And we are interested in finding the B/A division of the phasor, the resultant stationary vector will be

$R_{p}=\frac{B_{p}}{A_{p}}$

$R_{p}=\frac{\vert B\vert \angle30^{o}}{\vert A\vert \angle0^{o}}$

$R_{p}=\frac{\vert B\vert }{\vert A\vert } \angle(30-0)^{o}$

$R_{p}=\frac{\vert B\vert }{\vert A\vert } \angle30^{o}$

Sending